Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` of `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm Hg`. Determine the vapour pressure of `A` and `B` in their pure states.

Options

A. 500 and 600
B. 200 and 300
C. 300 and 400
D. 400 and 600

Answer

D

Explanation

Suppose X = A and Y = B
`P_(A)^(0)X_(A)+P_(B)^(0)X_(B)=P_("TOT")`
`therefore P_(A)^(0)((1)/(1+3))+P_(B)^(0)((3)/(1+3))=550 mm`
`therefore 0.25 P_(A)^(0)+0.75P_(B)^(0)=550 mm` ….(1)
Now add 1 mole B to mixture
`0.20 P_(A)^(0)+0.80 P_(B)^(0)=560 mm` .....(2)
multiply (1) by 4
`therefore P_(A)^(0)+3P_(B)^(0)=2200`
multiply (2) by 5
`therefore P_(A)^(0)+4P_(B)^(0)=2800`
`therefore P_(A)^(0)=400 mm =P_(X)^(0)`
`P_(A)^(0)=600 mm = P_(Y)^(0)`.